Car racing’s journey from rickety, steam-belching contraptions to today’s high-tech, world-famous supercars is packed with wild stories and weird traditions. It all kicked off in 1894 with the Paris-Rouen race, essentially a dare cooked up by a French newspaper to prove that these “horseless carriages” could actually go somewhere. Back then, “racing” was less about speed and more about keeping your vehicle in one piece — cars broke down, got lost, and generally limped to the finish. But people loved the spectacle, and car racing was born.
Soon, it wasn’t just local inventors; major companies joined in. In 1900, the Gordon Bennett Cup launched, inviting different nations to compete — the automotive Olympics of its time. National pride was on the line, and it was fierce. This competition laid the groundwork for some standard rules and introduced an international twist to the racing scene. Then, in 1906, we got the first true “Grand Prix” in France, where racers zoomed around closed circuits instead of weaving through the streets (and crowds), a move that saved some public panic and added to the event’s appeal.
With the 1910s came big names and epic distances, like the Indianapolis 500 in the U.S. This race gave drivers a chance to hammer down for 500 miles — at ridiculous speeds. In 1923, the 24 Hours of Le Mans took endurance to a new level, challenging racers to go non-stop for a full day and night. This legendary race tests not just speed but the mettle of both man and machine. By the time Formula One took off in 1950, car racing had become a global spectacle, with drivers achieving rockstar status and teams investing millions in shaving seconds off lap times.
Racetracks themselves became icons. The Monaco Grand Prix, weaving through city streets by the Mediterranean, is as glamorous as it is deadly, demanding precision and nerve. Silverstone, known as “the Home of British Motor Racing,” hosted the first Formula One race and has seen everything from spectacular wins to shocking crashes. The Nürburgring in Germany, nicknamed “The Green Hell” by legendary racer Jackie Stewart, is a track so twisty, challenging, and, frankly, terrifying that it’s in a league of its own.
In the 1960s and ’70s, racing entered a golden age of innovation, competition, and, yes, a little bit of craziness. The Ford-Ferrari rivalry became legendary as Ford took on Ferrari’s dominance at Le Mans. Formula One rivalries between teams like Ferrari, McLaren, and Lotus added drama on and off the track, fueling an era that left fans and media obsessed with the speed, danger, and glamour of racing.
Today, racing has grown beyond the screech of tires and roar of engines. Formula E is carving out a space for electric racing, while innovations in AI have introduced Roborace — racing, but with self-driving cars (think sci-fi meets Formula 1). Traditionalists may wince at the quiet hum of electric motors, but these changes are driving motorsport toward a greener, more futuristic path.
From rugged country roads to modern circuits packed with tech, car racing is a tale of competition, risk, and evolution — one that’s constantly fueled by innovation and a universal love for the thrill of speed.
My Sister at the Yas Marina Circuit
With the basics covered, we can now think more generally. Now we have to consider:
The weight of the car
The friction between the tires and the road
The radius of the osculating circle of the
turn
The bank angle
The Bristol Motor Speedway in Tennessee has two
straightaways and two turns. It is \(0.533\) miles long, with bank angles that
go all the way up to \(30^{\circ}\).
The track is mainly used for NASCAR race cars.
Let Babić’s race car of mass \(m\) with a constant
angular speed \(\omega\)
move around an arc of radius \(R\) banked at an angle of
\(\theta\). Its center of mass off the
ground by height \(h\) and
time \(t\) is measured in
seconds.
Let the car have a position function \(\vec{r}(t)\):
\(\qquad \vec{r}(t)=\begin{bmatrix} R\cos(\omega t) \\ R\sin(\omega t) \\ h \end{bmatrix}\)
To find the velocity function \(\vec{v}(t)\) of the car, we simply take the
derivative of \(\vec{r}(t)\) with
respect to \(t\):
\(\qquad \vec{v}(t)=\vec{r}^{\prime}(t) = \begin{bmatrix} -\omega R\sin(\omega t) \\ \omega R\cos(\omega t) \\ 0 \end{bmatrix}\)
Since the tangent line of a circle is
perpendicular to its radius, then if the
dot product of \(\vec{v}(t)\) and \(\vec{r}(t)\) is \(0\), then \(\vec{v}(t)\) is tangent to the
circular curve.
\(\qquad\begin{aligned} \vec{v}(t)\cdot\vec{r}(t) &= \begin{bmatrix} -\omega R\sin(\omega t) \\ \omega R\cos(\omega t) \\ 0 \end{bmatrix} \cdot \begin{bmatrix} R\cos(\omega t) \\ R\sin(\omega t) \\ h \end{bmatrix} \\~\\ &= -\omega R^2 \sin(\omega t) \cos(\omega t) + \omega R^2 \cos(\omega t) \sin(\omega t) \\~\\ &=0 \end{aligned}\)
Therefore the car’s velocity is tangent to the circular curve, and thus a centripetal force is needed to keep the car on the track.
To find speed \(v\), we
simply take the magnitude of the car’s velocity \(\vec{v}(t)\):
\(\qquad\begin{aligned} v &= \|\vec{v}(t)\| = \sqrt{\vec{v}(t)\cdot\vec{v}(t)} \\~\\ &= \sqrt{\omega ^2 R ^2 \left[\sin^2(\omega t) + \cos^2(\omega t)\right]} \\~\\ &= \sqrt{\omega ^2 R ^2} \\~\\ &= \omega R\end{aligned}\)
Since \(v=\omega R\), we can do some basic algebra do get the following equality:
\(\qquad\begin{aligned} &v=\omega R \\~\\ &\therefore \frac{1}{v}=\frac{1}{\omega R} \\~\\ &\therefore \frac{R}{v}=\frac{1}{\omega} \\~\\ &\therefore \frac{2\pi R}{v}=\frac{2\pi}{\omega} \end{aligned}\)
To find acceleration \(\vec{a}(t)\), we simply take the derivative
of the car’s velocity \(\vec{v}(t)\):
\(\qquad \vec{a}(t)=\vec{v}^{\prime}(t) = \begin{bmatrix} -\omega^2 R\cos(\omega t) \\ -\omega^2 R\sin(\omega t) \\ 0 \end{bmatrix}\)
Now to find the magnitude of acceleration \(a\):
\(\qquad\begin{aligned} a &= \|\vec{a}(t)\| = \sqrt{\vec{a}(t)\cdot\vec{a}(t)} \\~\\ &= \sqrt{\omega ^4 R ^2 \left[\cos^2(\omega t) + \sin^2(\omega t)\right]} \\~\\ &= \sqrt{\omega ^4 R ^2} \\~\\ &= \omega^2 R\end{aligned}\)
To see if \(\vec{a}(t)\) points
towards the center of the circle, we can take its
cross product with \(\vec{r}(t)\) with \(h=0\). This way, both \(\vec{a}(t)\) and \(\vec{r}(t)\) are on the \(xy-\text{plane}\), and the only way for
their cross product to not be \(\vec{0}\) is if they are not
parallel. Thus, if the cross product equals to \(\vec{0}\), then \(\vec{a}(t)\) is indeed
radial:
\(\qquad\begin{aligned} \vec{a}(t)\times\vec{r}(t) &= \begin{bmatrix} -\omega^2 R\cos(\omega t) \\ -\omega^2 R\sin(\omega t) \\ 0 \end{bmatrix} \times \begin{bmatrix} R\cos(\omega t) \\ R\sin(\omega t) \\ h=0 \end{bmatrix} \\~\\ &= \begin{bmatrix} 0 \\ 0 \\ -\omega^2R^2\cos(\omega t)\sin(\omega)t+\omega^2R^2\cos(\omega t)\sin(\omega)t \end{bmatrix} \\~\\ &= \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\~\\ &=\vec{0} \end{aligned}\)
Thus, the acceleration vector \(\vec{a}(t)\) points towards the center of the circle.
Let the centripetal force be \(\vec{F}_c(t)\) and its
magnitude be \(F_c\).
Since Newton's Second Law dictates that \(\vec{F}=m\vec{a}\), we have the following
three scalar equations:
\(\qquad\begin{cases} F_c=ma \\ a=\omega^2R \\ \frac{2\pi R}{v}=\frac{2\pi}{\omega} \end{cases}\)
To solve for \(F_c\) in terms of \(m\), \(v\), and \(R\), we can do some simple algebra:
\(\qquad\begin{aligned} &\frac{2\pi R}{v}=\frac{2\pi}{\omega} \\~\\ &\therefore \frac{R}{v}=\frac{1}{\omega} \\~\\ &\therefore \frac{v}{R}=\omega \end{aligned}\)
Now substitute \(\omega\) in Equation 2:
\(\qquad\begin{cases} F_c=ma \\ a=(\frac{v}{R})^2R=\frac{v^2}{R} \\ \frac{v}{R}=\omega \end{cases}\)
Now we can substitute \(a\) in Equation 1:
\(\qquad F_c=ma \therefore F_c=m\frac{v^2}{R}\)
The force that the ground exhibits on the car is perpendicular to the
track, and is called the normal force, denoted by \(\vec{N}\) with magnitude \(N\). As for friction, it is
denoted by \(\vec{f}\) with
magnitude \(f\). The
relationship between friction and the normal force is approximated by
this equation:
\(\qquad f=\mu N\)
\(\mu\) is a constant called the
coefficient of friction, and is dependent on the condition
of the track and the car’s tires.
The force of gravity can be described by \(mg\), where \(g\) is the
gravitational acceleration at the surface of the Earth,
which is around \(32.1740
\text{ft/s}^2\).
These three forces can be drawn on Babić race car like so:
We can abstract these forces and take their
vertical components:
To be precise, using basic trigonometry and
vector projection (proof can be found using
similar triangles), we can find that:
\(\qquad \begin{cases} N_z=N\cos\theta \\ f_z=f\sin\theta \\ (mg)_z=mg \end{cases}\)
Since the race car’s vertical acceleration (recall that
the z-component of \(\vec{a}(t)\) is always \(0\)), we know that all three of these
vertical components of forces must be sum to \(0\). Therefore, we can create an equation
(noting that the direction of the component corresponds to their
sign):
\(\qquad N_z-f_z-(mg)_z=0\)
With substitution, we can get:
\(\qquad N\cos\theta-f\sin\theta-mg=0\)
We can then rearrange to get:
\(\qquad N\cos\theta=f\sin\theta+mg\)
If we want to remove friction from the equation, then we
can use the equation \(f=\mu N\) to
substitute:
\(\qquad\begin{aligned} &N\cos\theta=f\sin\theta+mg \\~\\ &\therefore N\cos\theta=\mu N \sin \theta + mg \\~\\ &\therefore N(\cos\theta - \mu\sin\theta)=mg \\~\\ &\therefore N=\frac{mg}{\cos\theta - \mu\sin\theta} \end{aligned}\)
Similar to what we just did, instead of summing the vertical
components of the three forces, we can sum the horizontal
components.
Again, using trigonometry and
vector projection, we can find that:
\(\qquad \begin{cases} N_x=N\sin\theta \\ f_x=f\cos\theta \\ (mg)_x=0 \end{cases}\)
Because the magnitude of centripetal force \(F_c\) is just the sum of all the
horizontal components, we can list the following equation
and derive:
\(\qquad\begin{aligned} &F_c=N_x+f_x+(mg)_x &\therefore F_c=N\sin\theta+f\cos\theta\end{aligned}\)
But wait! We already have expressions for \(N\) and \(f\) in terms of \(\mu\). Let’s substitute that in:
\(\qquad\begin{aligned}&\begin{cases} F_c=N\sin\theta+f\cos\theta \\ f=\mu N \\ N=\frac{mg}{\cos\theta - \mu\sin\theta} \end{cases} \\~\\ &\therefore F_c=N\sin\theta+\mu N\cos\theta \\~\\ &\therefore F_c=N(\sin\theta+\mu\cos\theta) \\~\\ &\therefore F_c=\frac{mg}{\cos\theta-\mu\sin\theta}\cdot(\sin\theta+\mu\cos\theta) \\~\\ &\therefore F_c=\frac{\sin\theta+\mu\cos\theta}{\cos\theta-\mu\sin\theta}\cdot mg \end{aligned}\)
Now we have two equations for centripetal force:
\(\qquad\begin{cases} F_c=m\frac{v^2}{R} \\ F_c=\frac{\sin\theta+\mu\cos\theta}{\cos\theta-\mu\sin\theta}\cdot mg \end{cases}\)
Since second equation dictates the maximum centripetal
force the car can have, the first equation limits the maximum
speed the car can have before flying off the track. We can
substitute out \(F_c\) to find this
maximum speed \(v_{\text{max}}\):
\(\qquad\begin{aligned} &m\frac{v^2}{R}=\frac{\sin\theta+\mu\cos\theta}{\cos\theta-\mu\sin\theta}\cdot mg \\~\\ &\therefore v^2= \frac{\sin\theta+\mu\cos\theta}{\cos\theta-\mu\sin\theta}\cdot gR \end{aligned}\)
Because the mass of the car \(m\) is
not in the equation for \(v_{\text{max}}\), we can conclude that the
mass of the car actually does not affect the
maximum speed it can go through a turn with.
Let’s take a look back at the Bristol Motor Speedway.
The inside track of Turn 1 has a radius of \(211\text{ft}\) and a
bank angle of \(24^{\circ}\). If we assume dry
conditions so that the coefficient of friction \(\mu=0.98\), we can calculate the
maximum speed Babić’s race car can turn at:
\(\qquad\begin{aligned} v_{\text{max}}^2 &= \frac{\sin\theta+\mu\cos\theta}{\cos\theta-\mu\sin\theta}\cdot gR \\~\\ &= \frac{\sin24^{\circ}+0.98\cos24^{\circ}}{\cos24^{\circ}-0.98\sin24^{\circ}}\cdot 32\cdot211 \\~\\ &= 17072.1230 \end{aligned} \\~\\ \qquad \therefore v_{\text{max}}=\sqrt{17072.1230}=130.6603 \; (\text{ft/s})\)
But this is in feet per second. Let’s convert this to
miles per hour. Doing so, we can conclude that in dry
conditions, that maximum speed for a sharp turn is \(89.0866\text{mph}\).
In contrast, the outside track of Turn 1 has a radius of
\(251\text{ft}\) and a
bank angle of \(28^{\circ}\). If we still assume
dry conditions, we can calculate the
maximum speed Babić’s race car can turn at:
\(\qquad\begin{aligned} v_{\text{max}}^2 &= \frac{\sin\theta+\mu\cos\theta}{\cos\theta-\mu\sin\theta}\cdot gR \\~\\ &= \frac{\sin28^{\circ}+0.98\cos28^{\circ}}{\cos28^{\circ}-0.98\sin28^{\circ}}\cdot 32\cdot251 \\~\\ &= 25352.7302 \end{aligned} \\~\\ \qquad \therefore v_{\text{max}}=\sqrt{25352.7302}=159.2254 \; (\text{ft/s})\)
Converting this to miles per hour, we get that in dry
conditions, that maximum speed for a wide turn is \(108.5628\text{mph}\).
Back on the inside track of Turn 1, we’re now facing wet
conditions so that the coefficient of friction is now as
low as \(\mu=0.1\). Now, we calculate
that the maximum speed Babić’s race car can turn at:
\(\qquad\begin{aligned} v_{\text{max}}^2 &= \frac{\sin\theta+\mu\cos\theta}{\cos\theta-\mu\sin\theta}\cdot gR \\~\\ &= \frac{\sin24^{\circ}+0.1\cos24^{\circ}}{\cos24^{\circ}-0.1\sin24^{\circ}}\cdot 32\cdot211 \\~\\ &= 3852.9275 \end{aligned} \\~\\ \qquad \therefore v_{\text{max}}=\sqrt{3852.9275}=62.0720 \; (\text{ft/s})\)
Convert this to miles per hour, we can conclude that in
wet conditions, that maximum speed for a sharp turn is \(42.3218\text{mph}\).
Now let’s suppose that Babić’s race car is going around Turn 1 at
\(105\text{mph}\) (\(154\text{ft/s}\)) on the
outside track. We can then estimate her car’s tires’
coefficient of friction:
\(\qquad\begin{aligned} &v_{\text{max}}^2 = \frac{\sin\theta+\mu\cos\theta}{\cos\theta-\mu\sin\theta}\cdot gR \\~\\ &\therefore 154^2=\frac{\sin28^{\circ}+\mu\cos28^{\circ}}{\cos28^{\circ}-\mu\sin28^{\circ}}\cdot 32\cdot251 \\~\\ &\therefore 23716=\frac{0.4695+0.8829\mu}{0.8829-0.4695\mu}\cdot 8032 \\~\\ &\therefore \mu=\frac{17167.8324}{18226.1148}=0.9419 \end{aligned}\)
Thus, we can conclude that Babić’s car’s tires have a
coefficient of friction of \(\mu=0.9419\).
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